Chapter 5, Problem 13P

To determine

The ranking of the flywheels in descending order of the radial acceleration of a point on the rim.

Answer to Problem 13P

The ranking of the flywheels in descending order of the radial acceleration of a point on the rim is $\left(\text{c}\right),\left(\text{d}\right),\left(\text{a}\right),\left(\text{b}\right),\left(\text{e}\right)$ .

Explanation of Solution

Write the expression for the radial acceleration.

$a_r={\omega }^{2}r$ (I)

Here, $a_r$ is the radial acceleration, $w$ is the angular velocity and $r$ is the radius

Write the expression for the angular velocity.

$\omega =\frac{2\pi }{T}$

Here, $T$ is the period of the motion

Put the above equation in equation (I).

$\begin{array}{c}{a}_r={\left(\frac{2\pi }{T}\right)}^{2}r\\ a_r=4{\pi }^2\frac{r}{T^2}\\ a_r\propto \frac{r}{T^2}\end{array}$ (II)

Conclusion:

Find the values of $\left(\frac{r}{T^2}\right)$ for the different flywheels.

(a)

Given that the radius of the wheel is $8.0\text{ cm}$ and the period is $4.0\text{ ms}$ .

Substitute $8.0\text{ cm}$ for $r$ and $4.0\text{ ms}$ for $T$ in expression (II) to find $a_r$ .

$\begin{array}{c}{a}_r\propto \frac{8.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{{\left(4.0\text{ ms}\left(\frac{1\text{ s}}{1000\text{ ms}}\right)\right)}^2}\\ =\frac{0.08\text{ m}}{{\left(0.004\text{ s}\right)}^2}\\ =5000{\text{ m/s}}^2\end{array}$

(b)

Given that the radius of the wheel is $2.0\text{ cm}$ and the period is $4.0\text{ ms}$ .

Substitute $2.0\text{ cm}$ for $r$ and $4.0\text{ ms}$ for $T$ in expression (II) to find $a_r$ .

$\begin{array}{c}{a}_r\propto \frac{2.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{{\left(4.0\text{ ms}\left(\frac{1\text{ s}}{1000\text{ ms}}\right)\right)}^2}\\ =\frac{0.02\text{ m}}{{\left(0.004\text{ s}\right)}^2}\\ =1250{\text{ m/s}}^2\end{array}$

(c)

Given that the radius of the wheel is $8.0\text{ cm}$ and the period is $1.0\text{ ms}$ .

Substitute $8.0\text{ cm}$ for $r$ and $1.0\text{ ms}$ for $T$ in expression (II) to find $a_r$ .

$\begin{array}{c}{a}_r\propto \frac{8.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{{\left(1.0\text{ ms}\left(\frac{1\text{ s}}{1000\text{ ms}}\right)\right)}^2}\\ =\frac{0.08\text{ m}}{{\left(0.001\text{ s}\right)}^2}\\ =80000{\text{ m/s}}^2\end{array}$

(d)

Given that the radius of the wheel is $2.0\text{ cm}$ and the period is $1.0\text{ ms}$ .

Substitute $2.0\text{ cm}$ for $r$ and $1.0\text{ ms}$ for $T$ in expression (II) to find $a_r$ .

$\begin{array}{c}{a}_r\propto \frac{2.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{{\left(1.0\text{ ms}\left(\frac{1\text{ s}}{1000\text{ ms}}\right)\right)}^2}\\ =\frac{0.02\text{ m}}{{\left(0.001\text{ s}\right)}^2}\\ =20000{\text{ m/s}}^2\end{array}$

(e)

Given that the radius of the wheel is $1.0\text{ cm}$ and the period is $4.0\text{ ms}$ .

Substitute $1.0\text{ cm}$ for $r$ and $4.0\text{ ms}$ for $T$ in expression (II) to find $a_r$ .

$\begin{array}{c}{a}_r\propto \frac{1.0\text{ cm}\left(\frac{1\text{ m}}{100\text{ cm}}\right)}{{\left(4.0\text{ ms}\left(\frac{1\text{ s}}{1000\text{ ms}}\right)\right)}^2}\\ =\frac{0.01\text{ m}}{{\left(0.004\text{ s}\right)}^2}\\ =625{\text{ m/s}}^2\end{array}$

Therefore, the ranking of the flywheels in descending order of the radial acceleration of a point on the rim is $\left(\text{c}\right),\left(\text{d}\right),\left(\text{a}\right),\left(\text{b}\right),\left(\text{e}\right)$ .